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enter image description here

So this is what I got so far:

$$\int _0^{2\pi }\:\int _0^{\sqrt{5}}\:\int _3^{\sqrt{16-r^2}}\:\left(r\right)dzdrd\theta $$

And the answer is shown above in the screenshot. I'm not exactly sure which bounds I'm messing up.

I know it's a sphere, so it's gonna be $0$ to $2\pi$ for theta. The middle one I plug in $z=3$ and subtract and replace the squared $x$ and $y$ with $r$ squared. And then the furthest right one I do the same thing pretty much there.

What's my problem? Thank you.

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Perhaps it would be easier to start by finding the volume of a small slice of the sphere created by leaving only the portion of the sphere that lies in the range $z=a\dots b,\,b>a$.

This section would essentially be a flat pancake with its symmetry axis along z-axis. What is the radius of that pancake? Call it $\rho$. From basic Pythagoras theorem $\rho=\sqrt{r^2-z^2}$, where $r$ is the radius of the sphere. Clearly this only makes sense for $z^2\le r^2$.

So now you can find the volume of this pancake (note how positioning of $\int$ and $d\dots$ can be used to clarify the bounds in the integrals):

$$ V_{pancake}\left(a,b,r\right)=\int_a^b dz\int_0^\sqrt{r^2-z^2} \rho d\rho\int_0^{2\pi} d\phi=2\pi\int_a^b dz \frac{\left(\sqrt{r^2-z^2}\right)^2}{2}=\dots $$

From here it is easy to get the result for any portion of the sphere intersected by the plane

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  • $\begingroup$ woah that's kinda epic doe. what's a and b? $\endgroup$
    – Si Random
    Commented Oct 30, 2020 at 21:04
  • $\begingroup$ @Dr.SuessOfficial, it's at the beginning of my reply. Consider a sphere now remove the portion of the sphere below $z<a$ and above $z>b$. Provided $a$ and $b$ are close, you will be left with a flat pancake. Then you find the volume of that pancake. The final expression I gave above is valid even if the pancake is not thin, i.e. for any $-r\le a\le b\le r$ $\endgroup$
    – Cryo
    Commented Oct 30, 2020 at 21:55

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